Problem: Find $\lim_{h\to 0}\dfrac{6\arctan\left(1+h\right)-6\arctan\left(1\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $3$ (Choice C) C $6$ (Choice D) D The limit doesn't exist
Solution: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $6\arctan\left(1+h\right)-6\arctan\left(1\right)$, we can tell that the function is $f(x)=6\arctan(x)$ and the $x$ -value is $1$. In other words, the limit expression is equal to $f'\left(1\right)$ for $f(x)=6\arctan(x)$. Let's find $f'(x)$ : $f'(x)=6\cdot \dfrac{1}{{1+x^2}}$ Now let's evaluate $f'\left(1\right)$ : $\begin{aligned}f'\left(1\right)&=6\cdot \dfrac{1}{{1+\left(1\right)^2}} \\\\ &=3\end{aligned}$ In conclusion, $\lim_{h\to 0}\dfrac{6\arctan\left(1+h\right)-6\arctan\left(1\right)}{h}=3$.